Question

If you measured the final velocity of the car to be 1 m/s and the initial mass of the car remains the same, how much coal was added to the car in kg

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

`p = 14033.7 \ kg\cdot m/s`

b

`v = 1.22 \ m/s`

c

`m_2 = 9506.7 \ kg`

Step-by-step explanation:

From the question we are told that

The mass of the coal car is
`m_1 = 4527 \ kg`

The velocity is
`v_i = 3.1 \ m/s`

The load of coal dropped is
`m_2 = 6952 \ kg`

Generally the momentum of the system before the coal is added is mathematically represented as

`p = m_j * v_i`

=>
`p = 4527 * 3.1`

=>
`p = 14033.7 \ kg\cdot m/s`

Generally from the law of momentum conservation is mathematically represented as

`m_1 * v_i + m_2 * v_2 = (m_1 + m_2 )v`

Here
`v_2` is the velocity of the load before it is dropped and the value is 0 m/s

So

=>
`v =(m_1 * v_i )/(m_1 + m_2)`

`v = (4527 * 3.1 )/(( 4527 + 6952 ))`

=>
`v = 1.22 \ m/s`

When v = 1m/s and
`m_1` remain the same

Then

=>
`1 =(4527 * 3.1 )/(4527 + m_2)`

=>
`4527 + m_2 = 14033.7`

=>
`m_2 = 9506.7 \ kg`